Question 17 of 18intermediate🔧 ApplyNumerical4 marks

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65×104.65 \times 10⁻¹ N m⁻¹. The atmospheric pressure is 1.01×101.01 \times 10⁵ Pa. Also give the excess pressure inside the drop.

Correct Answer

Excess pressure inside the drop = 310 Pa (or 3.10×103.10 \times 10² Pa) Pressure inside the drop = 1.013×101.013 \times 10⁵ Pa

Exercise: EXERCISES | Q: 9.19 | (Page 22)
For More Understanding

Explanation

A mercury drop in air is a liquid drop surrounded by gas, having only one liquid surface. According to the textbook context, for such a case (similar to a bubble of gas in liquid), the excess pressure formula is 2S/r. The context explicitly states: 'The excess pressure in a bubble of gas in a liquid is given by 2S/r, where S is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case.' A mercury drop has one surface, unlike a soap bubble which has two surfaces (4S/r). The total pressure inside equals atmospheric pressure plus excess pressure.

Solution Steps

  1. Step 1: Convert radius to SI units. Radius r = 3.00 mm = 3.00×103.00 \times 10⁻³ m

  2. Step 2: Identify the formula for excess pressure. For a mercury drop (liquid drop in gas), there is one liquid surface, so excess pressure = 2S/r

  3. Step 3: Calculate the excess pressure. Excess pressure = 2S/r = 2 × (4.65×104.65 \times 10⁻¹) / (3.00×103.00 \times 10⁻³) = 0.93 / 0.003 = 310 Pa

  4. Step 4: Calculate the total pressure inside the drop. Total pressure = Atmospheric pressure + Excess pressure = 1.01×101.01 \times 10⁵ + 310 = 1.013×101.013 \times 10⁵ Pa

PreviousNext